0=-16t^2-20+36

Solution for 0=-16t^2-20+36 equation:

0=-16t^2-20+36

We move all terms to the left:

0-(-16t^2-20+36)=0

We add all the numbers together, and all the variables

-(-16t^2-20+36)=0

We get rid of parentheses

16t^2+20-36=0

We add all the numbers together, and all the variables

16t^2-16=0

a = 16; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·16·(-16)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*16}=\frac{32}{32}
=1
$